A hollow verticle cylinder of radius r and height h has a smooth internal surface. A small particle is placed in contact with the inner side of the upper rim, at point A, and given a horizontal speed u, tangential to the rim?

It leaves the lower rim at point B, vertically below A. If n is an integer then?

A) #u/(2pir)(sqrt((2h)/g)) = n#
B) #h/(2pir) = n#
C) #u/sqrt(2gh)=n#

Answer 1

I get
(A)

We see that vertical and horizontal speed of the particle are orthogonal and therefore can be treated independently. We also see that in all the three choices time #t# is not present and we are to find #n#
A . Vertical speed. The particle falls freely under action of gravity #g# It falls through height #h#. This is related by the kinematic expression
#s=ut+1/2at^2#

Inserting various values we get

#h=0xxt+1/2g t^2# #=>h=1/2g t^2# #=>h=1/2g t^2# #=>t=sqrt((2h)/g) ......(1)#
B. Horizontal speed. The particle is given horizontal speed #u#, tangential to the rim. As such its movement is around the walls of the cylinder as it rolls down in a spiral. The particle leaves the lower rim at point B, vertically below A. Therefore, at the time of exit it would have covered integral multiple #n# of its circumference. As such horizontal distance covered is #=n(2pir)#. Time taken for this #=t# The applicable expression is
#2npir=ut# #=>t=(2npir)/u# .......... (2)

Equating RHSs of (1) and (2) we get

#sqrt((2h)/g)=(2npir)/u# #=>n=(usqrt((2h)/g))/(2pir)#
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Answer 2

The motion of the particle is subject to the law of conservation of energy and conservation of angular momentum. Initially, the particle has kinetic energy due to its speed u and potential energy due to its height above the bottom of the cylinder. As it moves downward, its potential energy decreases, and its kinetic energy increases.

The angular momentum of the particle about the center of the cylinder is conserved because there is no net torque acting on it. Therefore, the angular momentum at the top (where it starts) is equal to the angular momentum at any point as it moves down.

By applying the conservation of energy and conservation of angular momentum, you can find equations that relate the particle's speed at any point to its distance from the bottom of the cylinder. These equations will help determine the particle's motion within the cylinder.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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